Question #78b35

1 Answer
Apr 25, 2017

"Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O"

Explanation:

Start by assigning oxidation numbers to the atoms that take part in the reaction

stackrel(color(blue)(-1))("Br")""^(-) + stackrel(color(blue)(+1))("H")_ 3 stackrel(color(blue)(+5))("As") stackrel(color(blue)(-2))("O")_ 4 -> stackrel(color(blue)(+5))("Br") stackrel(color(blue)(-2))("O")_ 3""^(-) + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+3))("As") stackrel(color(blue)(-2))("O")_ 2

Now, notice that the oxidation number of bromine is going from color(blue)(-1) on the reactants' side to color(blue)(+5) on the products' side. This implies that bromine is being oxidized, i.e. its oxidation number is increasing.

Similarly, the oxidation number of astatine if going from color(blue)(+5) on the reactants' side to color(blue)(+3) on the products' side. This implies that astatine is being reduced, i.e. its oxidation number is decreasing.

The oxidation half-reaction looks like this

stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-)

Now, the bromine atoms are balanced. To balance the oxygen atoms, you can use water molecules on the side that needs oxygen. In this case, you need 3 oxygen atoms on the reactants' side, so add 3 water molecules to that side of the equation

3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-)

To balance the hydrogen atoms, add protons, "H"^(+), to the side that needs hydrogen. In this case, you have 6 hydrogen atoms on the reactants' side, so add 6 protons to the products' side

3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)

Notice that the charge is balanced because you have

(1-) = (1-) + (6-) + (6+)

The reduction half-reaction looks like this

"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_2

Once again, the astatine atoms are balanced, so focus on the oxygen atoms first. You will have

"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O"

You now need 2 protons on the reactants' side to balance the hydrogen atoms

2"H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O"

Once again, notice that the charge is balanced because you have

(2+) + (2-) = 0

Now, in every redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you must multiply the reduction half-reaction by 3

{(color(white)(aaaaaaaa)3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)), (2"H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O" " " | xx 3) :}

to get

{(color(white)(aaaaaaaa)3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)), (6"H"^(+) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 6"e"^(-) -> 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + 6"H"_ 2"O") :}
color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)

3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) + color(red)(cancel(color(black)(6"H"^(+)))) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + color(red)(cancel(color(black)(6"e"^(-)))) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) + color(red)(cancel(color(black)(6"H"^(+)))) + 6"H"_2"O"

This will be equivalent to

color(darkgreen)(ul(color(black)("Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O")))

Therefore, you can say that water appears in the balanced chemical equation as a product with a coefficient of 3.

The bromide anion, "Br"^(-) is being oxidized to the bromate anion, "BrO"_3^(-), which implies that arsenic acid, "H"_3"AsO"_4, is acting as an oxidizing agent.

Consequently, you can say that arsenic acid is being reduced to arsonite, "HAsO"_2, which implies that the bromide anion is acting as a reducing agent.